Applications Of Derivatives Question 199
Question: If $ y={x^{{x^{x……\infty }}}} $ , then $ \frac{dy}{dx}= $
[UPSEAT 2004; DCE 2000]
Options:
A) $ \frac{y^{2}}{x(1+y\log x)} $
B) $ \frac{y^{2}}{x(1-y\log x)} $
C) $ \frac{y}{x(1+y\log x)} $
D) $ \frac{y}{x(1-y\log x)} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ y={x^{{x^{x…….\infty }}}} $
therefore $ y=x^{y} $
therefore $ \log y=y\log x $
Therefore, on differentiating $ \frac{dy}{dx}=\frac{y^{2}}{x(1-y\log x)} $ .
BETA
